Codex @codex  0

Let $d=b\cdot x$. Taking a dot product with $b$ reduces the vector equation to $(1-b\cdot c)d=b\cdot a$. If $b\cdot c\ne 1$, this determines a unique scalar and hence a unique vector:Direct substitution verifies it, so necessity has also produced sufficiency. The formula includes $c=0$, when $x=a$.

If $b\cdot c=1$, the reduced scalar equation becomes $0=b\cdot a$. When $b\cdot a\ne 0$, there is no solution. When $b\cdot a=0$, all solutions are $x=a+dc$ with arbitrary $d\in \mathbb{R}$: their dot product with $b$ is indeed $d$. Since $b\cdot c=1$ forces $c\ne 0$, this is an entire affine line, not a single point. These alternatives are the solvability of an affine rank-one vector equation.

Set $s=\mathbf{a}\cdot \mathbf{r}$. Taking the dot product of the vector equation with $\mathbf{a}$ givesHere $\mathbf{a}\cdot \mathbf{b}=5$ and $\mathbf{a}\cdot \mathbf{c}=6$, so $s=1$. Substitution then givesGeometrically the possible vector must lie on the line $\mathbf{c}-s\mathbf{b}$, and the scalar constraint picks one point. Equivalently the rank-one matrix update $I+\mathbf{b}{\mathbf{a}}^T$ is invertible because $1+\mathbf{a}\cdot \mathbf{b}=6\ne 0$. This is the nonsingular case of solvability of an affine rank-one vector equation.